find the domain of the function f(x)= 1/sqrt (4x-8) please show calculations

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

find the domain of the function f(x)= 1/sqrt (4x-8) please show calculations

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

huh
callum can you show the steps please
(2, infinity)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[f(x) = \frac{1}{\sqrt{4x-8}}\] now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need: \[4x-8 > 0\] \[x>\frac{1}{2}\] So I was wrong above, the domain is actually: \[\left(\frac{1}{2}, \infty \right)\]
x > 2
think you divided the wrong side there
oh yes sorry it's 2 lol, quite right. \[(2,\infty)\]
rickjbr can you demonstrate how you got your answer please
you'll have imaginary numbers if the square root is allowed to go negative
thus not part of the real number system
callum was right in his calculation 4x - 8 > 0 4(x-2) > 0 x-2 > 0 x > 2
yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2
thanks for the help to both of your
np
how about this. 2x^2-3x+5...find and simplify the difference quotient
The question reads Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5
ah, right
(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h
(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h
(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h
(2h^2+4xh+3x)/h
is that the final answer
checking, a bit rusty
my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h
i'll redo it, might've missed something
ok
ah, silly
that last step was it. instead of (2h^2+4xh+3x)/h it should be (2h^2+4xh-3h)/h
simply factor out an h
[h(2h+4x-3)]/h =2h+4x-3
that's it
thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum
probably better off posting on main forum, if i can help, i'll try
ok thanks for the help
Find \[\frac{f(x+h)-f(x)}{h}\] for \[f(x)=2x^2-3x+5\] \[\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}\] \[= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}\] \[=\frac{4xh+2h^2-3h}{h} = 4x+2h-3\] We can verify that this is right by noting that \[f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}\] The left hand side of this is equal to the derivative of f, which we can just do right here: \[f'(x) = 4x - 3\] and we should get that this equals the limit as h goes to zero of \[\frac{f(x+h) - f(x)}{h}.\] We have already calculated that \[\frac{f(x+h) - f(x)}{h} = 4x+2h-3\] so taking the limit of this as h goes to zero we get: \[\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.\] Hence we ave verified that the simplification has been done correctly.

Not the answer you are looking for?

Search for more explanations.

Ask your own question